 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2015/s7/1/3/solution

δ q_{i} = ε ξ_{i} (q, t)

, allow the Lagrangian to change by a total derivative,

δ L = ε d F / d t

. Direct variation gives

\frac{δ L}{ε} = \sum_{i} (\frac{\partial L}{\partial q _{i}} - \frac{d}{d t} \frac{\partial L}{\partial q ˙ _{i}}) ξ_{i} + \frac{d}{d t} \sum_{i} p_{i} ξ_{i} .

On a solution of the Euler-Lagrange equations, the first term vanishes. This proves Noether's theorem in the fixed-time form:

Q = i \sum p_{i} ξ_{i} - F, \dot{Q} = 0.

Spatial translations yield total momentum, and rotations yield angular momentum. Time translations yield energy when the Lagrangian has no explicit time dependence; this follows either from the extended time-transformation form of the theorem or by differentiating

H = \sum_{i} p_{i} \overset{q}{˙}_{i} - L

to obtain

\dot{H} = - \partial L / \partial t

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